Theore. For any positive integer n, 1 + 2 + ... + n = n(n+1)/2.
Proof. (Proof by Mathematical Induction) Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P(n) should be an assertion that for any n is verifiably either true or false.) The proof will now proceed in two steps: the initial step and the inductive step.
Initial Step. We must verify that P(1) is True. P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done with the initial step.
Inductive Step. Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true." Thus, we assume there is a k such that 1 + 2 + ... + k = k (k+1)/2. (We call this the inductive assumption.) We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.
This is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.
Since you have verified P(a), it follows from the inductive step that P(a+1) is true, and hence, P(a+2) is true, and hence P(a+3) is true, and so on. In this way the theorem has been proved.
Theorem. A formula for the sequence an defined above, is an = (1 - 1/22n)/2 for all n greater than or equal to 0.
Proof. (By Mathematical Induction.)
Initial Step. When n = 0, the formula gives us (1 - 1/22n)/2 = (1 - 1/2)/2 = 1/4 = a0. So the closed form formula ives us the correct answer when n = 0.
Inductive Step. Our inductive assumption is: Assume there is a k, greater than or equal to zero, such that ak = (1 - 1/22k)/2. We must prove the formula is true for n = k+1.
First we appeal to the recurrsive definition of ak+1 = 2 ak(1-ak). Next, we invoke the inductive assumption, for this k, to get
ak+1 = 2 (1 - 1/22k)/2 (1 - (1 - 1/22k)/2) = (1 - 1/22k)(1 + 1/22k)/2 = (1 - 1/22k+1)/2. This completes the inductive step.
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